Three number are in A.P, such that their sum is 18 and sum of there square is 158. The greatest among them is

Options

A. 10
B. 11
C. 12
D. none of these

Explanation

To solve the problem step-by-step, we need to find three numbers in arithmetic progression (A.P.) such that their sum is 18 and the sum of their squares is 158. Let's denote the three numbers as \( a - d \), \( a \), and \( a + d \), where \( a \) is the middle term and \( d \) is the common difference.

### Step 1: Set up the equations based on the problem statement.
1. The sum of the three numbers:
\[ (a - d) + a + (a + d) = 18 \]
Simplifying this, we get:
\[ 3a = 18 \]
Therefore,
\[ a = \frac{18}{3} = 6 \]

### Step 2: Use the sum of squares condition.
2. The sum of the squares of the three numbers:
\[ (a - d)^2 + a^2 + (a + d)^2 = 158 \]
Expanding this:
\[ (a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 158 \]
Combining like terms:
\[ 3a^2 + 2d^2 = 158 \]

### Step 3: Substitute the value of \( a \) into the equation.
3. Substitute \( a = 6 \) into the equation:
\[ 3(6^2) + 2d^2 = 158 \]
Calculating \( 6^2 \):
\[ 3(36) + 2d^2 = 158 \]
This simplifies to:
\[ 108 + 2d^2 = 158 \]

### Step 4: Solve for \( d^2 \).
4. Rearranging gives:
\[ 2d^2 = 158 - 108 \]
\[ 2d^2 = 50 \]
Dividing by 2:
\[ d^2 = 25 \]
Taking the square root:
\[ d = 5 \quad \text{or} \quad d = -5 \]

### Step 5: Find the three numbers.
5. Now we can find the three numbers:
- If \( d = 5 \):
\[ a - d = 6 - 5 = 1, \quad a = 6, \quad a + d = 6 + 5 = 11 \]
- If \( d = -5 \):
\[ a - d = 6 - (-5) = 11, \quad a = 6, \quad a + d = 6 + (-5) = 1 \]

### Step 6: Identify the greatest number.
6. The three numbers are \( 1, 6, 11 \) (or \( 11, 6, 1 \) depending on the value of \( d \)). The greatest among them is:
\[ \text{Greatest number} = 11 \]

### Final Answer:
The greatest among the three numbers is **11**.

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